Session3: Correlation and Linear Regression

• correlation

• linear regression

A common task in statistical analysis is to investigate the linear relationship between pairs of numeric vectors.

This can be done by identifying the correlation between numeric vectors using the cor() function in R.

In this example we use cor() to identify the Pearson correlation between two variables. The method argument may be set to make use of different correlation methods.

• Perfectly posively correlated vectors will return 1
• Perfectly negatively correlated vectors will return -1
• Vectors showing no or little linear correlation will be close to 0.

Often we wish to apply correlation analysis to all columns or rows in a matrix in a pair-wise manner. To do this in R, we can simply pass the cor() function a single argument of the numeric matrix of interest. The cor() function will then perform all pair-wise correlations between columns.

• subset iris dataset

> data(iris)
> head(iris)

  Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1          5.1         3.5          1.4         0.2  setosa
2          4.9         3.0          1.4         0.2  setosa
3          4.7         3.2          1.3         0.2  setosa
4          4.6         3.1          1.5         0.2  setosa
5          5.0         3.6          1.4         0.2  setosa
6          5.4         3.9          1.7         0.4  setosa

> iris4cor<-iris[,c(1:4)]; 
> # or
> # iris4cor<-iris[,-5]

> pairs(iris4cor)

We have seen how we can find the linear correlation between two sets of variables using cor() function.

R also provides a comprehensive set of tools for regression analysis including the well used linear modeling function lm()

• least square method

minimize the vertical distance between the fitted line and data points

The lm() function fits a linear regression to your data and provides useful information on the generated fit.

In the example below we fit a linear model using lm() on the kid.weights dataset with weight (Y) as the dependent variable and height (X) as the explanatory variable.

> lmResult<-lm(formula = weight ~ height, data = kid.weights)
> lmResult

Call:
lm(formula = weight ~ height, data = kid.weights)

Coefficients:
(Intercept)       height  
    -31.342        1.909  

As we have seen, printing the model result provides the intercept and slope of line. To get some more information on the model we can use the summary() function

> summary(lmResult)

Call:
lm(formula = weight ~ height, data = kid.weights)

Residuals:
    Min      1Q  Median      3Q     Max 
-24.837  -7.361  -2.384   3.616  68.708 

Coefficients:
             Estimate Std. Error t value Pr(>|t|)    
(Intercept) -31.34191    3.17493  -9.872   <2e-16 ***
height        1.90904    0.08343  22.882   <2e-16 ***
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Residual standard error: 14.09 on 248 degrees of freedom
Multiple R-squared:  0.6786,    Adjusted R-squared:  0.6773 
F-statistic: 523.6 on 1 and 248 DF,  p-value: < 2.2e-16

Predict the kid weight with the height information.

If we have 3 kids with height = 23, 40 and 66 inches, how do we predict their weights?

Use the information from the $coefficients

> new_kid_height<-c(23,40,66)
> beta0<-lmResult$coefficients[1]
> beta1<-lmResult$coefficients[2]
> 
> predicted_new_kid_weight<-beta0+beta1*new_kid_height
> predicted_new_kid_weight

[1] 12.56609 45.01984 94.65497

Or use the predict()

new_kid_height_df <- data.frame(height=c(23,40,66))
cleaver_predicted_kid_weight<-predict(lmResult,new_kid_height_df)
cleaver_predicted_kid_weight

       1        2        3 
12.56609 45.01984 94.65497 

The residuals are the difference between the predicted and actual values. To retrieve the residuals we can access the slot or use the resid() function.

> summary(resid(lmResult))

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
-24.840  -7.361  -2.384   0.000   3.616  68.710 

> summary(lmResult$residual)

   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
-24.840  -7.361  -2.384   0.000   3.616  68.710 

Ideally you would want your residuals to be normally distributed around 0.

\[ E[e_{i}]=0 \]

Plot the residuals

plot(kid.weights$height,kid.weights$weight,ylim=c(-30,150),
     ylab="weigth (pounds)",xlab="height (inches)")
abline(lmResult,col="blueviolet",lwd=3, lty=1)

Residual is the vertical distance between the observed data and the regression line. It has the same unit as the dependent variable.

Plot the residuals against the independent variable (X), i.e. the height. It makes the residual accessment easiler by eyes.

plot(kid.weights$height,lmResult$residual,ylim=c(-30,150),
     ylab="residuals (weight pounds)",xlab="height (inches)")
abline(h=0,col="blueviolet",lwd=3, lty=1)

Plot the residuals against the independent variable (X), i.e. the height.

• The R-squared value represents the proportion of variability in the response variable that is explained by the explanatory variable.

• A high R-squared here indicates that the line fits closely to the data.

> summary(lmResult)$r.squared

[1] 0.6785746

• Question: How would you describe (or summarize) kid's weight when the height information is absence?

> kid.weights$weight

  [1]  38  87  50  98  47  30  24  45 144  24  40  15  20  10  19  29  80
 [18]  20  28  24  35  30  52  72  28  14  18  11  31  28  31  14  38  15
 [35]  27  60  40  40  21  47  46  30  69  76  16  11  38  25  29  21  36
 [52]  28  41  13  21  33  40  32  27  20  30  45  34  19  41  40  19  14
 [69]  43  90  45  29 105  94  22  69  17  27  12  40  25  13  20  16  45
 [86]  27  20  54  60  30  18  55  14  34  45  48  30  25  38  13  22  19
[103]  74  30  89  35  33  78  43  40  32  31  15  19  55  40  37  34  45
[120]  44  50  50  28  32  22  52  40  42  22  23  16  17  30  40  32  26
[137]  30  61  65  52  23  29  18  32  50  13  36  47  80  23  64  45  25
[154]  65  26  38  20 150  42  53  35  16 150  25  30  52  27  27  22  30
[171]  35  42  12  47  14  32  25  93  30  59 100  34  52  43  32  13  18
[188]  38  17  46  29  50  14  13  26  11  32  70  16  50  26  23  37  60
[205]  45  26 131  22  12  70  86  35  14  85  58  28  17  34  19  13  18
[222]  23  17  30  41  30  13 125  32  18  35  32 108  40  42  28  26  26
[239]  23  42  28  55  65  80  49  18  14  26  85  14

• Question: How would you describe (or summarize) kid's weight when the height information is absence?

• mean might be a good choice

> mean(kid.weights$weight)

[1] 38.384

• Question: How would you describe (or summarize) kid's weight when the height information is absence?

• mean might be a good choice

The fraction of variability in the independent variable (Y; or the weight in this example) that can be explained by the explanatory variable (X; or the height in this example).

\[ TSS=\text{Total Sum of Squares}=\sum_{i=1}^n(y_i-\overline y)^2 \\ SSE=\text{Sum of the Square of the residuals}=\sum_{i=1}^n(y_i-\hat{y})^2 \]

> SSE<-sum(lmResult$residuals^2)
> TSS<-sum((kid.weights$weight - mean(kid.weights$weight))^2)
> R_square<-1-(SSE/TSS)
> R_square

[1] 0.6785746

> summary(lmResult)$r.squared

[1] 0.6785746

The results from linear models also provides a measure of significance for a variable not being relevant

> summary(lmResult)$fstatistic

   value    numdf    dendf 
523.5631   1.0000 248.0000 

\[ F=\frac{MSM}{MSE}=\frac{\text{mean of the explained variance}}{\text{mean of the unexplained variance}}=\frac{({\displaystyle\frac{SSM}1})}{({\displaystyle\frac{SSE}{n-2}})} \]

> n=nrow(kid.weights)
> SSM <- sum((predict(lmResult) - mean(kid.weights$weight))^2)
> MSE <-sum(lmResult$residuals^2)/(n-2)
> 
> MSM <-SSM/1
> 
> MSM/MSE

[1] 523.5631

> summary(lmResult)$fstatistic

   value    numdf    dendf 
523.5631   1.0000 248.0000 

Exercise on this session can be found here

Answers can be found here

R code for solutions can be found here [here]